# Implicit Differentiation: How Chain Rule is applied vs. Explicit Differentiation

Please assume this is my initial exposure to calculus and derivatives.

I am having difficulty making the connection between the application of the chain rule to explicit differentiation and that of implicit differentiation. Everything I’ve learned so far about differentiation has been based on explicitly defined functions and limits. Applying the chain rule to explicit functions makes sense to me, as I am just recognizing composite functions within an original function. But applying the chain rule to the dependent variable in an implicit function doesn't feel the same to me. I know it works, but it just doesn't feel right. Take for example the equation $x\sqrt{y}=1$. I understand that $y$ is a function of $x$, but it is the very function $x\sqrt{y}=1$ that defines the relationship between x and y. This is different than how I have applied the chain rule when performing explicit differentiation. Take for example the function $y=x\sin(2x^2+2x+1)$. Recognizing that $2x^2+2x+1$ is a composite function, and applying the chain rule, makes sense to me. But for some reason, I cannot wrap my head around making that same association to $y$ in the function $x\sqrt{y}=1$.

Perhaps I am not completely understanding the chain rule. What am I missing? Thanks for any help you can provide.

• Would you have the same conceptual difficulties if both $x$ and $y$ in $x\sqrt y$ were functions of $t$ and you were differentiating with respect to $t$? – amd Feb 22 '18 at 0:22
• No one said that x and y were not related in implicit differentiation, just that it is highly unlikely that for general combinations of x and y that you are able to explicitly express one variable as a clear function of the other. – Triatticus Feb 22 '18 at 2:10
• @amd, You’ve got me thinking. I would not have the same conceptual difficulty if both $x$ and $y$ were functions of $t$. If I think about $\frac{d}{dx}$ as an operator, then differentiating both sides of an equation does not affect the balance of the equation. For example, if I take the simple equation $y=x$, square both sides, and then take the derivative, the answer must be the same as if I took the derivative of the $y=x$ equation. Is thinking about $\frac{d}{dx}$ as an operator acceptable? – slax Feb 22 '18 at 17:48
• Certainly! That point of view will be very useful when you study differential equations. Note, though, that in the process of applying an operator to both sides you might introduce spurious solutions, as is the case with your example of squaring both sides. – amd Feb 22 '18 at 20:07
• Thank you for the clarification. – slax Feb 23 '18 at 1:04

Consider: $g(x)=x\sqrt{y(x)}$. Then: $$g'(x)=(x\sqrt{y(x)})'=1\cdot \sqrt{y(x)}+x\cdot \frac{1}{2\sqrt{y(x)}}\cdot y'(x)=0 \Rightarrow \\ y'(x)=\frac{-\sqrt{y(x)}\cdot 2\sqrt{y(x)}}{x}=-\frac{2y(x)}{x}.$$

Addendum: You are right that the balance must be preserved when $\frac{d}{dx}$ is applied: $$\frac{d}{dx}\left(x\sqrt{y}\right)=\frac{d}{dx}(1) \Rightarrow \sqrt{y}\frac{d}{dx}(x)+x\frac{d}{dx}(\sqrt{y})=0 \Rightarrow \cdots.$$ Also the differential of a function is $dy=y'dx$. Hence: $$y=x \Rightarrow dy=dx \Rightarrow y'dx=dx \Rightarrow y'=1 \\ y^2=x^2 \Rightarrow dy^2=dx^2 \Rightarrow 2ydy=2xdx \Rightarrow 2y\cdot y'dx=2xdx \Rightarrow y'=\frac{x}{y}=\frac{x}{x}=1.$$

• I know I am not explaining my issue well. My issue is that $y(x)=\frac{1}{x^2}$, which is obtained by solving the equation being differentiated, seems different than recognizing that $2x^2+2x+1$ is a composite function of $y=x\sin(2x^2+2x+1)$. I think I may have a resolution. If I think about $\frac{d}{dx}$ as an operator, then differentiating both sides of an equation must not affect the balance of the equation. For example, take the simple equation $y=x$, square both sides, and then take the derivative. $[y^2]$ must be differentiated with respect to $y$ to keep the equation balanced. – slax Feb 22 '18 at 18:44
• Thank you for the addendum. Is it correct to say that $\frac{d}{dx}[y^2]=\frac{d}{dx}[x^2] \Rightarrow \frac{d}{dy}[y^2]\cdot\frac{d}{dx}[y]=\frac{d}{dx}[x^2] \cdot\frac{d}{dx}[x] \Rightarrow 2y\cdot\frac{dy}{dx}=2x\cdot\frac{dx}{dx} \Rightarrow 2y\cdot\frac{dy}{dx}=2x\cdot1\Rightarrow \frac{dy}{dx}=\frac{x}{y}$ – slax Feb 23 '18 at 0:58
• Yes, correct. However, note that for $xy^2=1$, it is $\frac{d}{dx}(xy^2)=\frac{d}{dx}(1) \Rightarrow y^2\cdot \frac{d}{dx}(x)+x\cdot \frac{d}{dx}(y^2)=y^2+x\cdot \frac{d}{dy}(y^2)\cdot \frac{d}{dx}(y)=0.$ – farruhota Feb 23 '18 at 3:28

For the function

$$x\sqrt{y}=1\implies x=\frac1{\sqrt y}\implies \frac{dx}{dy}=-\frac12\frac1{y\sqrt y}$$

or as an alternative by implicit differentiation

$$x\sqrt{y}=1\implies dx \sqrt y+x\frac1{2\sqrt y}dy=0\implies dx \sqrt y=-x\frac1{2\sqrt y}dy \\\implies\frac{dx}{dy}=-\frac{x}{2y}=-\frac12\frac1{y\sqrt y}$$